elementul X ionizeaza prin acceptare de e si formeaza ionul negativ Xⁿˉ
12,8g X...........4,8176
*10²³ e
A g..............n*6,023*10²³
e ==> 32*4,8176**10²³ =12,8*n* 6,023*10²³ (1)
dar:
A g................ 6,023*10²³
atomi
53,12*10
ˉ²⁴
g ........1
atom ==> A=32
in relatia (1) inlocuiesti A=32 si afli n= 1,99 aproxim. n=2 X²ˉ